# How to make a Median absolute deviation of $N(0, \sigma^2)$ an unbiased estimator of $\sigma$, asymptotically?

I am looking for a derivation of the fact that $$\frac{1}{\Phi^{-1}(3/4)}$$ is the multiplier needed for the Median Absolute Deviation (MAD) to be an unbiased estimator of $$\sigma$$ when $$x_i\sim N(0, \sigma^2)$$. Recall the MAD is defined as:

$$\lambda = b\times Median(| X-Median(X)|).$$

for some $$b$$ chosen to meet a given criteria, (e.g. often unbiased-ness) the claim is stated in many places (wikipedia and journal articles) but I cannot find a proof.

• Are you satisfied that it is sufficient to show that $\frac{\text{MAD}}{\sigma}$ at the normal is $\Phi^{-1}(3/4)$? Commented Feb 10, 2020 at 3:50
• @Glen_b-ReinstateMonica yeah that would be equivalent Commented Feb 10, 2020 at 3:57

An outline of why it is the case that $$\frac{\text{MAD}}{\sigma}$$ at the normal is $$\Phi^{-1}(3/4)$$ (where here $$\text{MAD}$$ is the median absolute deviation from the median).
For the normal, the population median is at $$\mu$$.
Then consider the distribution of $$\frac{X-\mu}{\sigma}$$. This is just a standard normal. Hence $$\frac{|X-\mu|}{\sigma}$$ is the absolute value of a standard normal. Its median will be the value $$m$$ such that the standard normal has half the area between $$-m$$ and $$m$$. We already know that half the distribution is between the first and third quartile, so $$m$$ is the upper quartile.
Consequently in a general normal distribution, the ratio $$\frac{\text{MAD}}{\sigma}$$ is the upper quartile, $$\Phi^{-1}(3/4)$$.
Now for the normal (and indeed in other symmetric distributions with finite mean with positive density at the median) the sample median is consistent as an estimator of $$\mu$$, so $$X-\tilde{x}$$ will asymptotically go to $$X-\mu$$, and so in turn asymptotically the median of $$|X-\tilde{x}|$$ will be the median of $$|X-\mu|$$.
As a result, to get a consistent estimator of $$\sigma$$ from the MAD (median absolute deviation from the median) for a random sample from a normal distribution, you divide the MAD by $$\Phi^{-1}(3/4)$$.